HDU 2647 Reward 反向拓补排序

最新推荐文章于 2021-07-19 16:34:02 发布

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最新推荐文章于 2021-07-19 16:34:02 发布

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本文介绍了一个基于拓扑排序的算法，用于解决工厂老板如何以最少的资金满足员工间相对工资要求的问题。每个员工的工资需至少为888，并且满足某些员工要求其工资高于特定同事的条件。

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Reward

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2824 Accepted Submission(s): 830

Problem Description

Dandelion's uncle is a boss of a factory. As the spring festival is coming , he wants to distribute rewards to his workers. Now he has a trouble about how to distribute the rewards.
The workers will compare their rewards ,and some one may have demands of the distributing of rewards ,just like a's reward should more than b's.Dandelion's unclue wants to fulfill all the demands, of course ,he wants to use the least money.Every work's reward will be at least 888 , because it's a lucky number.

Input

One line with two integers n and m ,stands for the number of works and the number of demands .(n<=10000,m<=20000)
then m lines ,each line contains two integers a and b ,stands for a's reward should be more than b's.

Output

For every case ,print the least money dandelion 's uncle needs to distribute .If it's impossible to fulfill all the works' demands ,print -1.

Sample Input

Sample Output

  
   1777
-1

Author

dandelion

Source

曾是惊鸿照影来

Recommend

yifenfei

题意是说老板发工资了，员工a要求的工资要比员工b多，每个员工最少工资为888，老板至少发多少钱能够满足每个员工的要求。

因为要求a要比b多，所以在拓补排序的时候要将b作为入度，a作为出度，然后用邻接表来存不容易超内存，最后直接拓补一下就可以。

#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
#define M 10007
int head[M],in[M],money[M];
int n,m;

struct Edge
{
    int to,next;
}edg[M*2];

void init()
{
    for(int i=1;i<=n;i++)
    {
        in[i]=0;
        head[i]=-1;
        money[i]=888;
    }
}

int topsort()
{
    queue<int>q;
    int ans=0,num=0;
    for(int i=1;i<=n;i++)//将一开始入度为0的所有点加入队列
        if(in[i]==0)
            q.push(i);
    while(!q.empty())
    {
        int u=q.front();
        ans+=money[u];
        q.pop();
        num++;
        for(int k=head[u];k!=-1;k=edg[k].next)
            if(--in[edg[k].to]==0)//找到要求比他工资高的人
            {
                money[edg[k].to]=money[u]+1;//将此人工资在其要求的人工资基础加1
                q.push(edg[k].to);
            }
    }
    if(num!=n)//如果所求的员工数与所给的员工数不同，则返回-1
        return -1;
    else return ans;
}

int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        init();
        int a,b,k=0;
        while(m--)
        {
            scanf("%d%d",&a,&b);
            in[a]++;
            edg[k].to=a;//以a为出度
            edg[k].next=head[b];
            head[b]=k++;
        }
        printf("%d\n",topsort());
    }
    return 0;
}